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Basic Concepts of Probability (2.2)
Copyright © 2009 Pearson Education, Inc.
Slide 1- 1
Dealing with Random Phenomena
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A random phenomenon is a situation in which we know what
outcomes could happen, but we don’t know which particular
outcome did or will happen.
In general, each occasion upon which we observe a random
phenomenon is called a trial.
At each trial, we note the value of the random phenomenon, and
call it an outcome.
When we combine outcomes, the resulting combination is an
event.
The collection of all possible outcomes is called the
sample space.
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Slide 1- 2
The Law of Large Numbers
First a definition . . .
 When thinking about what happens with
combinations of outcomes, things are simplified if
the individual trials are independent.
 Roughly speaking, this means that the outcome
of one trial doesn’t influence or change the
outcome of another.
 For example, coin flips are independent.
Copyright © 2009 Pearson Education, Inc.
Slide 1- 3
The Law of Large Numbers (cont.)

The Law of Large Numbers (LLN) says that the
long-run relative frequency of repeated
independent events gets closer and closer to a
single value.

We call the single value the probability of the
event.
Because this definition is based on repeatedly
observing the event’s outcome, this definition of
probability is often called empirical probability.

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Slide 1- 4
The Nonexistent Law of Averages
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The LLN says nothing about short-run behavior.
Relative frequencies even out only in the long run, and
this long run is really long (infinitely long, in fact).
The so called Law of Averages (that an outcome of a
random event that hasn’t occurred in many trials is
“due” to occur) doesn’t exist at all.
Copyright © 2009 Pearson Education, Inc.
Slide 1- 5
Foundation of Probability
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The onset of probability as a useful science is
primarily attributed to Blaise Pascal (1623-1662) and
Pierre de Fermat (1601-1665). While contemplating
a gambling problem posed by Chevalier de Mere in
1654, Blaise Pascal and Pierre de Fermat laid the
fundamental groundwork of probability theory, and
are thereby accredited the fathers of probability.
Chances of a Lifetime
(16 minutes in)
Copyright © 2009 Pearson Education, Inc.
Slide 1- 6
Modeling Probability (cont.)

The probability of an event is the number of
outcomes in the event divided by the total number
of possible outcomes.
P(A) =
# of outcomes in A
# of possible outcomes
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Slide 1- 7
Example
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A bag contains 5 red, 4 blue, 3 green and 8 orange
marbles.
What is the probability of selecting a green marble?
Copyright © 2009 Pearson Education, Inc.
Slide 1- 8
Formal Probability
1. Two requirements for a probability:

A probability is a number between 0 and 1.
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For any event A, 0 ≤ P(A) ≤ 1.
2. Probability Assignment Rule:
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The probability of the set of all possible
outcomes of a trial must be 1.

P(S) = 1 (S represents the set of all possible
outcomes.)
Copyright © 2009 Pearson Education, Inc.
Slide 1- 9
Formal Probability (cont.)
3. Complement Rule:
 The set of outcomes that are not in the event
A is called the complement of A, denoted AC.
 The probability of an event occurring is 1
minus the probability that it doesn’t occur:
P(A) = 1 – P(AC)
Slide 1- 10
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Example


A bag contains 5 red, 4 blue, 3 green and 8 orange
marbles.
What is the probability of selecting a green, blue or red
marble?
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Slide 1- 11
Formal Probability
5. Multiplication Rule:

For two independent events A and B, the
probability that both A and B occur is the
product of the probabilities of the two events.

P(A and B) = P(A) x P(B), provided that A
and B are independent.
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Slide 1- 12
Formal Probability (cont.)
4. Addition Rule:

Events that have no outcomes in common
(and, thus, cannot occur together) are called
disjoint (or mutually exclusive).
Slide 1- 13
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Formal Probability (cont.)
4. Addition Rule (cont.):

For two disjoint events A and B, the
probability that one or the other occurs is the
sum of the probabilities of the two events.

P(A or B) = P(A) + P(B), provided that A and
B are disjoint.
Slide 1- 14
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Example
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Using the whole numbers from 1 – 20.
What is the probability of selecting a number less than
7 OR greater than 15?
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Slide 1- 15
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General Addition Rule:
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For any two events A and B,
P(A or B) = P(A) + P(B) – P(A and B)

(On the Formula Sheet)
The following Venn diagram
shows a situation in which we
would use the
general addition rule:
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Slide 1- 16
Conditional Probabilities

To find the probability of the event B given the
event A, we restrict our attention to the outcomes
in A. We then find in what fraction of those
outcomes B also occurred.
P
(A
and
B)
P(B|A) 
P(A)

(On the Formula Sheet)
Note: P(A) cannot equal 0, since we know that A
has occurred.
Copyright © 2009 Pearson Education, Inc.
Slide 1- 17
The General Multiplication Rule

When two events A and B are
independent, we can use the multiplication
rule for independent events:
P(A and B) = P(A) x P(B)

However, when our events are not
independent, this earlier multiplication rule
does not work. Thus, we need the General
Multiplication Rule.
Copyright © 2009 Pearson Education, Inc.
Slide 1- 18
The General Multiplication Rule (cont.)
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We encountered the general multiplication rule in
the form of conditional probability.
Rearranging the equation in the definition for
conditional probability, we get the General
Multiplication Rule:
 For any two events A and B,
P(A and B) = P(A) x P(B|A)
or
P(A and B) = P(B) x P(A|B)
Copyright © 2009 Pearson Education, Inc.
Slide 1- 19
What are Tree Diagrams
• A way of showing the possibilities of two or
more events
• Simple diagram we use to calculate the
probabilities of two or more events
For example – a fair coin is spun twice
1st
2nd
H
HH
T
HT
H
TH
T
TT
H
T
Possible
Outcomes
Attach probabilities
1st
½
½
2nd
½
H
H
HH
P(H,H)=½x½=¼
T
HT
P(H,T)=½x½=¼
½
H
TH
P(T,H)=½x½=¼
½
T
TT
P(T,T)=½x½=¼
½
T
INDEPENDENT EVENTS – 1st spin has no effect on the 2nd spin
Calculate probabilities
1st
½
½
2nd
½
H
H
HH
P(H,H)=½x½=¼
*
T
HT
P(H,T)=½x½=¼
½
H
TH
P(T,H)=½x½=¼
*
*
½
T
TT
P(T,T)=½x½=¼
½
T
Probability of at least one Head?
For example – 10 coloured beads in a bag – 3 Red, 2 Blue,
5 Green. One taken, colour noted, returned to bag, then a
second taken.
1st
2nd
R
B
G
R
RR
B
RB
G
R
RG
BR
B
BB
G
R
BG
GR
B
GB
G
GG
INDEPENDENT EVENTS
Probabilities
1st
2nd
0.3
0.2
0.3
R
0.5
0.3
0.2
0.2
B
0.5
0.5
0.3
0.2
G
0.5
R
RR
P(RR) = 0.3x0.3 = 0.09
B
RB
P(RB) = 0.3x0.2 = 0.06
G
R
RG
BR
P(RG) = 0.3x0.5 = 0.15
P(BR) = 0.2x0.3 = 0.06
B
BB
P(BB) = 0.2x0.2 = 0.04
G
R
BG
GR
P(BG) = 0.2x0.5 = 0.10
P(GR) = 0.5x0.3 = 0.15
B
GB
P(GB) = 0.5x0.2 = 0.10
G
GG
P(GG) = 0.5x0.5 = 0.25
All ADD UP to 1.0
Choose a meal
Main course
Salad 0.2
Egg & Chips 0.5
Pizza 0.3
IC
Pudding
Ice Cream 0.45
Apple Pie 0.55
P(S,IC) = 0.2 x 0.45 = 0.09
AP
P(S,AP) = 0.2 x 0.55 = 0.110
0.45
IC
P(E,IC) = 0.5 x 0.45 = 0.225
0.55
AP
P(E,AP) = 0.5 x 0.55 = 0.275
IC
P(P,IC) = 0.3 x 0.45 = 0.135
AP
P(P,AP) = 0.3 x 0.55 = 0.165
0.45
S
0.55
0.2
0.5
E
0.3
P
0.45
0.55
Basic Counting Rule
Example: If you have 5 Shirts, 4 Pairs of Pants and 3 Pairs of
Shoes. You can make (5)(4)(3) = 60 different outfits.
Factorial
• Factorial means you multiply by all the
whole numbers less than that number down
to one.
Permutations
• Example: I have 9 paintings and have room to display
4 of them at a time on my wall. How many different
ways can I do this?
Combinations
Definition
• Let’s say that a game gives payoffs
a1, a2,…, an with probabilities
p1, p2,… pn.
* The expected value (or expectation)
E of this game is
E = a1p1 + a2p2 + … + anpn.
• Think of expected value as a long term
average.
American Roulette
• At a roulette table in Las Vegas, you will find the following
numbers 1 – 36, 0, 00. There are 38 total numbers.
• We place a $1 chip on 7. If the ball lands in the 7 slot we win
$35 (net winnings). If the ball lands on any other number we
lose our $1 chip.
• What is the expectation of this bet?
• To answer this question we need to know the probability of
winning and losing.
• The probability of winning is 1/38. The probability of losing is
37/38.
• So the expectation is E = $35(1/38) + (-$1)(37/38) =
(35-37)/38 = -2/38 = -$0.053
• What this tells us is that over a long time for every $1 we bet we
will lose $0.053.
• This is an example of a game with a negative expectation. One
should not play games when the expectation is negative.
Example 2
• On the basis of previous experience a librarian knows
that the number of books checked out by a person
visiting the library has the following probabilities:
# of books
Probability
0
1
0.15 0.35
2
3
4
5
0.25
0.15
0.05
0.05
• Find the expected number of books checked out
by a person.
• E = 0(0.15)+1(0.35)+2(0.25)+3(0.15)+4(0.05)+5(0.05)
• E = 0 + 0.35 + 0.50 + 0.45 + 0.20 + 0.25
• E = 1.75